Proof We prove only the first statement about the source map s. The second statement about the range map r follows in a similar way. Using the definitions of $A_s$ and permutation matrices, one verifies that

$$\begin{align*}A_s P_\tau = P_\sigma A_s \iff {(A_s)}_{v\tau(e)} = {(A_s)}_{\sigma^{-1}(v) e} \quad \forall v \in V, \, e \in E, \end{align*}$$

where the right side is equivalent to

$$\begin{align*}v \in s(\tau(e)) \iff \sigma^{-1}(v) \in s(e) \qquad \forall v \in V, \, e \in E. \end{align*}$$

However, $\sigma ^{-1}(v) \in s(e)$ can be rewritten as $v \in \sigma (s(e))$ , so that we obtain

$$\begin{align*}v \in s(\tau(e)) \iff v \in \sigma(s(e)) \qquad \forall v \in V, \, e \in E. \end{align*}$$

This means that $s(\tau (e)) = \sigma (s(e))$ for all $e \in E$ .

Proof Let $\Gamma := (V, E)$ and denote by $\mathcal {A}$ the $C^*$ -algebra $C(\operatorname {\mathrm {Aut}}^+(\Gamma ))$ . Then $\operatorname {\mathrm {Spec}} \mathcal {A}$ is a group with multiplication given by $\varphi * \psi := (\varphi \otimes \psi ) \circ \Delta $ and is isomorphic to a subgroup of unitary matrices $G \subseteq \mathbb {C}^{(V \sqcup E) \times (V \sqcup E)}$ via

$$\begin{align*}\varphi \in \operatorname{\mathrm{Spec}} \mathcal{A} \ \longleftrightarrow \ \varphi(u) := (\varphi(u_{ij})) \in \mathbb{C}^{(V \sqcup E) \times (V \sqcup E)}, \end{align*}$$

see, for example, [Reference Timmermann31, Proposition 6.1.11]. Further, we have the decomposition $u = u_E \oplus u_V$ , so that $\varphi (u) = \varphi (u_E) \oplus \varphi (u_V)$ is given by a pair of matrices $\varphi (u_E)$ and $\varphi (u_V)$ . Since $u_V$ and $u_E$ are magic unitaries, $\varphi (u_E)$ and $\varphi (u_V)$ are exactly permutation matrices, which correspond to a pair of permutations $(\sigma , \tau ) \in S_V \times S_E$ via the permutation representation from Definition 3.2. Proposition 3.3 then implies that $(\sigma , \tau )$ are exactly automorphisms of $\Gamma $ .

Proof Denote by u the fundamental representation of $\operatorname {\mathrm {Aut}}^+(\Gamma )$ and by $\widehat {u}$ the fundamental representation of $S_V^+ * S_E^+$ . Since $u_V$ and $u_E$ are magic unitaries, the universal property of $C(S_V^+ * S_E^+)$ implies the existence of a unital $*$ -homomorphism $\varphi \colon C(S_V^+ * S_E^+) \to C(\operatorname {\mathrm {Aut}}^+(\Gamma ))$ with $\varphi (\widehat {u}_{vw}) = u_{vw}$ and $\varphi (\widehat {u}_{ef}) = u_{ef}$ . This $*$ -homomorphism is surjective since $C(\operatorname {\mathrm {Aut}}^+(\Gamma ))$ is generated by the entries of $u_V$ and $u_E$ . Further, one verifies directly that it is a morphism of compact quantum groups. Thus, $\operatorname {\mathrm {Aut}}^+(\Gamma ) \subseteq S_V^+ * S_E^+$ .

Proof Denote by u the fundamental representation of $\operatorname {\mathrm {Aut}}^+(\Gamma _{n,m})$ and by $\widehat {u}$ the fundamental representation of $S_V^+ * S_E^+$ . By the proof of Proposition 3.9, we have $\operatorname {\mathrm {Aut}}^+(\Gamma _{n,m}) \subseteq S_V^+ * S_E^+$ via a unital $*$ -homomorphism $\varphi \colon C(S_V^+ * S_E^+) \to C(\operatorname {\mathrm {Aut}}^+(\Gamma _{n,m}))$ with $\varphi (\widehat {u}_{vw}) = u_{vw}$ and $\varphi (\widehat {u}_{ef}) = u_{ef}$ . To show the other inclusion, we construct the inverse $*$ -homomorphism using the universal property of $C(\operatorname {\mathrm {Aut}}^+(\Gamma _{n,m}))$ . Thus, we have to show that $\widehat {u}_V$ and $\widehat {u}_E$ satisfy the relations from Definition 3.4. However, $\widehat {u}_V$ and $\widehat {u}_{E}$ are magic unitaries by definition, and we compute

$$\begin{align*}\sum_{\substack{f \in E \\ v \in s(f)}} \widehat{u}_{fe} = \sum_{f \in E} \widehat{u}_{fe} = 1 = \sum_{w \in V} \widehat{u}_{vw} = \sum_{\substack{w \in V \\ w \in s(e)}} \widehat{u}_{vw} \quad \forall v \in V, e \in E, \end{align*}$$

so that $A_s \widehat {u}_E = \widehat {u}_V A_s$ by Remark 3.6. Similarly, one shows that $A_r \widehat {u}_E = \widehat {u}_V A_r$ by replacing the source map s with the range map r. Thus, the $*$ -homomorphism from Proposition 3.9 is invertible, which shows $\operatorname {\mathrm {Aut}}^+(\Gamma _{n,m}) = S_V^+ * S_E^+$ .

Proof Let $\Gamma := (V, E)$ . Denote by u the fundamental representation of $\operatorname {\mathrm {Aut}}^+(\Gamma )$ and by $\widehat {u}$ the fundamental representation of $\operatorname {\mathrm {Aut}}^+(\Gamma ^{\operatorname {\mathrm {op}}})$ . By definition, we have $A_{s^{\operatorname {\mathrm {op}}}} = A_r$ and $A_{r^{\operatorname {\mathrm {op}}}} = A_s$ , so that the entries of u and $\widehat {u}$ satisfy exactly the same relations. Hence, by the universal properties of $C(\operatorname {\mathrm {Aut}}^+(\Gamma ))$ and $C(\operatorname {\mathrm {Aut}}^+(\Gamma ^{\operatorname {\mathrm {op}}}))$ , there exists a $*$ -isomorphism $\varphi \colon C(\operatorname {\mathrm {Aut}}^+(\Gamma )) \to C(\operatorname {\mathrm {Aut}}^+(\Gamma ^{\operatorname {\mathrm {op}}}))$ with $\varphi (u_{vw}) = \widehat {u}_{vw}$ and $\varphi (u_{ef}) = \widehat {u}_{ef}$ . Further, a direct computation shows that this $*$ -isomorphism is indeed a morphism of compact quantum groups.

Proof Let $\Gamma := (V, E)$ and denote by u the fundamental representation of $\operatorname {\mathrm {Aut}}^+(\Gamma )$ and by $\widehat {u}$ the fundamental representation of $\operatorname {\mathrm {Aut}}^+(\Gamma ^*)$ . We begin by constructing a $*$ -isomorphism $\varphi \colon C(\operatorname {\mathrm {Aut}}^+(\Gamma )) \to C(\operatorname {\mathrm {Aut}}^+(\Gamma ^*))$ with $\varphi (u_{vw}) = \widehat {u}_{vw}$ and $\varphi (u_{ef}) = \widehat {u}_{ef}$ using the universal properties of $C(\operatorname {\mathrm {Aut}}^+(\Gamma ))$ and $C(\operatorname {\mathrm {Aut}}^+(\Gamma ^*))$ . Hence, we have to show that the entries of u and $\widehat {u}$ satisfy the same relations. Since $u_V$ , $u_E$ , $\widehat {u}_E$ , and $\widehat {u}_V$ are all magic unitaries, it only remains to show that

$$\begin{align*}A_s u^{(1)} = u^{(2)} A_s \iff A_{s^*} u^{(2)} = u^{(1)} A_{s^*}, \end{align*}$$

$$\begin{align*}A_r u^{(1)} = u^{(2)} A_r \iff A_{r^*} u^{(2)} = u^{(1)} A_{r^*} \end{align*}$$

for two arbitrary magic unitaries $u^{(1)}$ and $u^{(2)}$ indexed by V and $E,$ respectively. However, this follows directly from Proposition 2.24, since

$$\begin{align*}{(A_{s^*})}_{ev} = \begin{cases} 1 & \text{if } e \in s^*(v), \\ 0 & \text{otherwise}, \end{cases} = \begin{cases} 1 & \text{if } v \in s(e), \\ 0 & \text{otherwise}, \end{cases} = {(A_s)}_{ve} = {(A_s^T)}_{ev} \end{align*}$$

for all $e \in E$ and $v \in V$ , so that $A_{s^*} = A_s^T = A_s^*$ and similarly $A_{r^*} = A_r^*$ . Thus, the $*$ -isomorphism $\varphi $ exists and one verifies directly that it is a morphism of compact quantum groups.

Proof Let $e \in E$ and $X \subseteq V$ . Fix an arbitrary ordering $X = \{v_1, \ldots , v_k \}$ , where all $v_i$ are distinct. Then

$$\begin{align*}\prod_{v \in X} {(A_s u_E)}_{v e} = \prod_{v \in X} \sum_{\substack{f \in E \\ v \in s(f)}} u_{fe} = \sum_{\substack{f_1 \in E \\ v_1 \in s(f_1)}} \dots \sum_{\substack{f_k \in E \\ v_k \in s(f_k)}} u_{f_1 e} \dots u_{f_k e}. \end{align*}$$

Since $u_{f_1 e} \dots u_{f_k e} = \delta _{f_1 f_2} \dots \delta _{f_1 f_k} u_{f_1 e}$ , it follows that

$$\begin{align*}\prod_{v \in X} {(A_s u_E)}_{v e} = \sum_{\substack{f \in E \\ v_1 \in s(f), \ldots, v_k \in s(f)}} u_{f e} = \sum_{\substack{f \in E \\ X \subseteq s(f)}} u_{fe}. \end{align*}$$

On the other hand, we can apply $A_s u_E = u_V A_s$ to the original expression to obtain

$$\begin{align*}\prod_{v \in X} {(A_s u_E)}_{v e} = \prod_{v \in X} {(u_V A_s)}_{v e} = \prod_{v \in X} \sum_{\substack{w \in V \\ w \in s(e)}} u_{vw}. \end{align*}$$

Thus,

$$\begin{align*}\sum_{\substack{f \in E \\ X \subseteq s(f)}} u_{fe} = \prod_{v \in X} \sum_{\substack{w \in V \\ w \in s(e)}} u_{vw}. \end{align*}$$

Since the left side does not depend on the order of X, the product on the right side commutes. Further, we can replace s with r to obtain a proof of the corresponding statement for the range map r.

Proof By rewriting the statement using Lemma 3.14, we have to show that

$$\begin{align*}\sum_{\substack{f \in E \\ s(f) = X}} u_{fe} = \sum_{X \subseteq Y \subseteq V} {(-1)}^{\left\lvert Y\right\rvert - \left\lvert X\right\rvert} \sum_{\substack{f \in E \\ Y \subseteq s(f)}} u_{fe} \quad \forall e \in E. \end{align*}$$

Now, consider an element $u_{fe}$ with $X \subseteq s(f)$ and define $k := |s(f)| - |X|$ . Then there are exactly $\binom {k}{\ell }$ subsets Y with $X \subseteq Y \subseteq s(f)$ and $|X| + \ell $ elements. Further, each is weighted with a factor of

$$\begin{align*}{(-1)}^{|Y|-|X|} = {(-1)}^{(|X| + \ell) - |X|} = {(-1)}^{\ell} \end{align*}$$

on the right side of the equation. Thus, by the binomial theorem, the total contribution of $u_{ef}$ on the right side is

$$\begin{align*}\sum_{\ell = 0}^k \binom{k}{\ell} {(-1)}^{\ell} = {((-1) + 1)}^k = \begin{cases} 1 & \text{if } k = 0, \\ 0 & \text{if } k> 0. \end{cases} \end{align*}$$

Therefore, the right side of the equation contains exactly each $u_{fe}$ with $k = 0$ , which is equivalent to $s(f) = X$ . The corresponding statement for the range map r can be proven in the same way.

Proof Let $e, f \in E$ . Then

$$\begin{align*}\sum_{\substack{g \in E \\ s(g) = s(e)}} u_{gf}, \sum_{\substack{g \in E \\ r(g) = r(e)}} u_{gf} \in C^*(u_V) \end{align*}$$

by choosing $X = s(e)$ and $X = r(e)$ in Lemma 3.15. This implies

$$\begin{align*}\bigg( \sum_{\substack{g \in E \\ s(g) = s(e)}} u_{gf} \bigg) \bigg( \sum_{\substack{g \in E \\ r(g) = r(e)}} u_{gf} \bigg) = \sum_{\substack{g_1, g_2 \in E \\ s(g_1) = s(e) \\ r(g_2) = r(e)}} \underbrace{u_{g_1 f} u_{g_2 f}}_{\delta_{g_1 g_2} u_{g_1 f}} = \sum_{\substack{g \in E \\ s(g) = s(e) \\ r(g) = r(e)}} u_{gf} \in C^*(u_V). \end{align*}$$

However, $\Gamma $ has no multi-edges, so that

$$\begin{align*}u_{ef} = \sum_{\substack{g \in E \\ s(g) = s(e) \\ r(g) = r(e)}} u_{gf} \in C^*(u_V). \end{align*}$$

Proof Denote by $\textit {u}$ the fundamental representation of $\operatorname {\mathrm {Aut}}^+(\Gamma )$ and by $\widehat {u}$ the fundamental representation of $S_V^+$ . As in the proof of Proposition 3.9, there exists a morphism of compact quantum groups $\varphi \colon C(S_V^+) \to \operatorname {\mathrm {Aut}}^+(\Gamma )$ with $\varphi (\widehat {u}_{vw}) = u_{vw}$ . By Theorem 3.16, this morphism is surjective, so that $\operatorname {\mathrm {Aut}}^+(\Gamma ) \subseteq S_V^+$ .

Proof By combining Proposition 3.4 with Corollary 3.17, we obtain $\operatorname {\mathrm {Aut}}^+(\Gamma ) = \operatorname {\mathrm {Aut}}^+(\Gamma ^*) \subseteq S_E^+$ .

Proof Let $v_0 \in V$ , $e := (v_1, w_1) \in E$ . Then

$$\begin{align*}{(A_s u_E)}_{v_0 e} = \sum_{\substack{f \in E \\ v_0 \in s(f)}} u_{fe} = \sum_{\substack{(v_2, w_2) \in E \\ v_0 = v_2}} u_{(v_2, w_2)(v_1, w_1)}. \end{align*}$$

On the other hand, the image of s always contains one element and there are no multi-edges so that

$$\begin{align*}{(u_V A_s)}_{v_0 e} = \sum_{\substack{w \in V \\ w \in s(e)}} u_{v_0 w} = u_{v_0 v_1}. \end{align*}$$

Thus, $A_s u_E = u_V A_s$ is equivalent to

$$\begin{align*}\sum_{\substack{(v_2, w_2) \in E \\ v_0 = v_2}} u_{(v_2, w_2)(v_1, w_1)} = u_{v_0 v_1} \qquad \forall v_0 \in V, \, (v_1, w_1) \in E. \end{align*}$$

The equivalence between $A_r u_E = u_V A_r$ and the second equation in the statement follows similarly.

Proof Let $(v_1, w_1), (v_2, w_2) \in E$ . By Lemma 4.1, we have

$$\begin{align*}u_{v_1 v_2} = \sum_{\substack{(v_3, w_3) \in E \\ v_1 = v_3}} u_{(v_3, w_3)(v_2, w_2)}, \qquad u_{w_1 w_2} = \sum_{\substack{(v_4, w_4) \in E \\ w_1 = w_4}} u_{(v_4, w_4)(v_2, w_2)}, \end{align*}$$

which yields

$$ \begin{align*} u_{v_1 v_2} u_{w_1 w_2} &= \sum_{\substack{(v_3, w_3) \in E \\ v_1 = v_3}} \sum_{\substack{(v_4, w_4) \in E \\ w_1 = w_4}} \underbrace{u_{(v_3, w_3)(v_2, w_2)} u_{(v_4, w_4)(v_2, w_2)}}_{\delta_{(v_3, w_3)(v_4, w_4)} u_{(v_3, w_3)(v_2, w_2)}} = \sum_{\substack{(v_3, w_3) \in E \\ v_1 = v_3 \\ w_1 = w_3}} u_{(v_3, w_3)(v_2, w_2)}, \\ u_{w_1 w_2} u_{v_1 v_2} &= \sum_{\substack{(v_4, w_4) \in E \\ w_1 = w_4}} \sum_{\substack{(v_3, w_3) \in E \\ v_1 = v_3}} \underbrace{ u_{(v_4, w_4)(v_2, w_2)} u_{(v_3, w_3)(v_2, w_2)}}_{\delta_{(v_4, w_4)(v_3, w_3)} u_{(v_3, w_3)(v_2, w_2)}} = \sum_{\substack{(v_3, w_3) \in E \\ v_1 = v_3 \\ w_1 = w_3}} u_{(v_3, w_3)(v_2, w_2)}. \end{align*} $$

Since $\Gamma $ has no multi-edges, we have

$$\begin{align*}u_{(v_1, w_1) (v_2, w_2)} = \sum_{\substack{(v_3, w_3) \in E \\ v_1 = v_3 \\ w_1 = w_3}} u_{(v_3, w_3)(v_2, w_2)}. \end{align*}$$

Thus,

$$\begin{align*}u_{(v_1, w_1) (v_2, w_2)} = u_{v_1 v_2} u_{w_1 w_2} = u_{w_1 w_2} u_{v_1 v_2}. \end{align*}$$

Proof Let $\Gamma := (V, E)$ . Denote by u the fundamental representation of $\operatorname {\mathrm {Aut}}^+(\Gamma )$ and by $\widehat {u}$ the fundamental representation of $\operatorname {\mathrm {Aut}}^+_{\operatorname {\mathrm {Bic}}}(\Gamma )$ . Further, define the elements $\widehat {u}_{(v_1, w_1)(v_2, w_2)} := \widehat {u}_{v_1 v_2} \widehat {u}_{w_1 w_2}$ for all $(v_1, w_1), (v_2, w_2) \in E$ , and the matrices $\widehat {u}_V := {(\widehat {u}_{vw})}_{v,w \in V}$ and $\widehat {u}_E := {(\widehat {u}_{ef})}_{e,f \in E}$ .

To prove the statement, we begin by constructing a unital $*$ -homomorphism $\varphi \colon C(\operatorname {\mathrm {Aut}}^+(\Gamma )) \to C(\operatorname {\mathrm {Aut}}^+_{\operatorname {\mathrm {Bic}}}(\Gamma ))$ defined by $\varphi (u_{vw}) = \widehat {u}_{vw}$ and $\varphi (u_{(v_1,w_1)(v_2,w_2)}) = \widehat {u}_{v_1 v_2}\widehat {u}_{w_1 w_2}$ using the universal property of $C(\operatorname {\mathrm {Aut}}^+(\Gamma ))$ . Thus, we have to show that the matrices $\widehat {u}_V$ and $\widehat {u}_E$ satisfy the relations from Definition 3.4. By Definition 2.27, $\widehat {u}_V$ is a magic unitary, and we have

$$\begin{align*}\widehat{u}_{v_1 v_2} \widehat{u}_{w_1 w_2} = \widehat{u}_{w_1 w_2} \widehat{u}_{v_1 v_2} \quad \forall (v_1, w_1),(v_2, w_2) \in E, \end{align*}$$

which implies

$$ \begin{align*} {(\widehat{u}_{v_1 v_2} \widehat{u}_{w_1 w_2})}^* &= {(\widehat{u}_{w_1 w_2})}^* {(\widehat{u}_{v_1 v_2})}^* = \widehat{u}_{w_1 w_2} \widehat{u}_{v_1 v_2} = \widehat{u}_{v_1 v_2} \widehat{u}_{w_1 w_1}, \\ {(\widehat{u}_{v_1 v_2} \widehat{u}_{w_1 w_2})}^2 &= \widehat{u}_{v_1 v_2} \widehat{u}_{w_1 w_2} \widehat{u}_{v_1 v_2} \widehat{u}_{w_1 w_2} = {(\widehat{u}_{v_1 v_2})}^2 {(\widehat{u}_{w_1 w_2})}^2 = \widehat{u}_{v_1 v_2} \widehat{u}_{w_1 w_2}. \end{align*} $$

Further, the additional relations from Proposition 2.29 yield

$$ \begin{align*} & \sum_{(v_2, w_2) \in E} \widehat{u}_{v_1 v_2} \widehat{u}_{w_1 w_2} = \sum_{\substack{v_2, w_2 \in V \\ (v_2, w_2) \in E}} \widehat{u}_{v_1 v_2} \widehat{u}_{w_1 w_2} = 1, \\ & \sum_{(v_2, w_2) \in E} \widehat{u}_{v_2 v_1} \widehat{u}_{w_2 w_1} = \sum_{\substack{v_2, w_2 \in V \\ (v_2, w_2) \in E}} \widehat{u}_{v_2 v_1} \widehat{u}_{w_2 w_1} = 1 \end{align*} $$

for all $(v_1, w_1) \in E$ . Hence, $\widehat {u}_E$ is a magic unitary. Next, we check the intertwiner relation $A_s \widehat {u}_E = \widehat {u}_V A_s$ . Let $v_0 \in V$ and $(v_1, w_1) \in E$ . Then

$$\begin{align*}&\sum_{\substack{(v_2, w_2) \in E \\ v_0 = v_2}} \widehat{u}_{v_2 v_1} \widehat{u}_{w_2 w_1} = \widehat{u}_{v_0 v_1} \sum_{\substack{(v_2, w_2) \in E \\ v_0 = v_2}} \widehat{u}_{w_2 w_1} = \widehat{u}_{v_0 v_1} \sum_{w_2 \in V} {(A_{\Gamma})}_{v_0 w_2 }\widehat{u}_{w_2 w_1}. \end{align*}$$

By Definition 2.27, we have $A_{\Gamma } \widehat {u}_V = \widehat {u}_V A_{\Gamma }$ , so that

$$\begin{align*}\widehat{u}_{v_0 v_1} \sum_{w_2 \in V} {(A_{\Gamma})}_{v_0 w_2 }\widehat{u}_{w_2 w_1} = \widehat{u}_{v_0 v_1} \sum_{w_2 \in V} \widehat{u}_{v_0 w_2 } {(A_{\Gamma})}_{w_2 w_1}. \end{align*}$$

Since $\widehat {u}_{v_0 v_1} \widehat {u}_{v_0 w_2} = \delta _{v_1 w_2} \widehat {u}_{v_0 v_1}$ , it follows that

$$\begin{align*}\widehat{u}_{v_0 v_1} \sum_{w_2 \in V} \widehat{u}_{v_0 w_2 } {(A_{\Gamma})}_{w_2 w_1} = \sum_{w_2 \in V} \delta_{v_1 w_2} \widehat{u}_{v_0 v_1} {(A_{\Gamma})}_{w_2 w_1} = \widehat{u}_{v_0 v_1} {(A_{\Gamma})}_{v_1 w_1} = \widehat{u}_{v_0 v_1}. \end{align*}$$

Thus, we have in total

$$\begin{align*}\sum_{\substack{(v_2, w_2) \in E \\ v_0 = v_2}} \widehat{u}_{(v_2, w_2)(v_1, w_1)} = \sum_{\substack{(v_2, w_2) \in E \\ v_0 = v_2}} \widehat{u}_{v_2 v_1} \widehat{u}_{w_2 w_1} = \widehat{u}_{v_0 v_1} , \end{align*}$$

which is equivalent to $A_s \widehat {u}_E = \widehat {u}_V A_s$ by Lemma 4.1. Similarly, one shows $A_r \widehat {u}_E = \widehat {u}_V A_r$ so that the map $\varphi $ exists.

Next, we construct the inverse map $\psi \colon C(\operatorname {\mathrm {Aut}}^+_{\operatorname {\mathrm {Bic}}}(\Gamma )) \to C(\operatorname {\mathrm {Aut}}^+(\Gamma ))$ given by $\psi (\widehat {u}_{vw}) = u_{vw}$ using the universal property of $C(\operatorname {\mathrm {Aut}}^+_{\operatorname {\mathrm {Bic}}}(\Gamma ))$ . Thus, we have to show that the matrix $u_V$ satisfies the relations from Definition 2.27. First, note that $u_V$ is a magic unitary by Definition 3.4. Second, we have to show that $A_{\Gamma } u_V = u_V A_{\Gamma }$ . Observe that $A_{\Gamma } = A_s A_r^*$ , since

$$\begin{align*}{(A_s A_r^*)}_{v w} = \sum_{e \in E} {(A_s)}_{v e} {(A_r)}_{w e} = \sum_{e \in E} \delta_{(v, w)e} = \begin{cases} 1 & \text{if } (v, w) \in E, \\ 0 & \text{otherwise}, \end{cases} \end{align*}$$

for all $v, w \in V$ . Hence,

$$\begin{align*}A_{\Gamma} u_V = A_s A_r^* u_V = A_s u_E A_r^* = u_V A_s A_r^* = u_V A_{\Gamma}, \end{align*}$$

because $A_r^*$ also intertwines $u_V$ and $u_E$ by Remark 3.6. Finally, we have to check that $u_{v_1 v_2} u_{w_1 w_2} = u_{w_1 w_2} u_{v_1 v_2}$ for all edges $(v_1, w_1), (v_2, w_2) \in E$ . But this follows directly from Lemma 4.2, since

$$\begin{align*}u_{v_1 v_2} u_{w_1 w_2} = u_{(v_1, w_1)(v_2, w_2)} = u_{w_1 w_2} u_{v_1 v_2}. \end{align*}$$

Thus, the $*$ -homomorphism $\psi $ exists. Note that the maps $\varphi $ and $\psi $ are indeed inverse, because

$$\begin{align*}\widehat{u}_{v w} \ \longleftrightarrow \ u_{v w}, \qquad \widehat{u}_{v_1 v_2}\widehat{u}_{w_1 w_2} \ \longleftrightarrow \ u_{v_1 v_2} u_{w_1 w_2} = u_{(v_1, w_1)(v_2, w_2)} \end{align*}$$

by Lemma 4.2. Further, a direct computation shows that they are morphisms of compact quantum groups.

Proof Since $A_s = A_r$ , we only have to consider $A_s u_E = u_V A_s$ . Let $v_0 \in V$ and $e := \{ v_1, w_1 \} \in E$ . Then

$$ \begin{align*} {(A_s u_E)}_{v_0 e} &= \sum_{f \in E} {(A_s)}_{v_0 f} u_{f e} = \sum_{\substack{f \in E \\ v_0 \in f}} u_{f e} = \sum_{\substack{v_2 \in V \\ \{ v_0, v_2 \} \in E }} u_{\{v_0, v_2\} \{ v_1, w_1 \}}, \\ {(u_V A_s)}_{v_0 e} &= \sum_{w \in V} u_{v_0 w} {(A_s)}_{w e} = \sum_{\substack{w \in V \\ w \in e}} u_{v_0 w} = u_{v_0 v_1} + u_{v_0 w_1}. \end{align*} $$

Hence, $A_s u_E = u_V A_s$ is equivalent to

$$\begin{align*}\sum_{\substack{v_2 \in V \\ \{ v_0, v_2 \} \in E }} u_{\{v_0, v_2\} \{ v_1, w_1 \}} = u_{v_0 v_1} + u_{v_0 w_1} \quad \forall v_0 \in V, \, \{v_1, w_1\} \in E. \end{align*}$$

Similarly, we compute

$$ \begin{align*} {(u_E A_s^*)}_{e v_0} &= \sum_{f \in E} u_{e f} {(A_s^*)}_{f v_0} = \sum_{\substack{f \in E \\ v_0 \in f}} u_{e f} = \sum_{\substack{v_2 \in V \\ \{ v_0, v_2 \} \in E }} u_{\{ v_1, w_1 \} \{v_0, v_2\}}, \\ {(A_s^* u_V )}_{e v_0} &= \sum_{w \in V} {(A_s)}_{e w} u_{w v_0} = \sum_{\substack{w \in V \\ w \in e}} u_{w v_0} = u_{v_1 v_0} + u_{w_1 v_0}. \end{align*} $$

Thus, $u_E A_s^* = A_s^* u_V$ is equivalent to

$$\begin{align*}\sum_{\substack{v_2 \in V \\ \{ v_0, v_2 \} \in E }} u_{\{ v_1, w_1 \} \{v_0, v_2\}} = u_{v_1 v_0} + u_{w_1 v_0} \quad \forall v_0 \in V, \, \{v_1, w_1\} \in E, \end{align*}$$

which is again equivalent to $A_s u_E = u_V A_s$ by Proposition 2.24.

Proof Since $v_1 \neq w_1$ , we have

$$ \begin{align*} (u_{v_1 v_2} + u_{v_1 w_2})(u_{w_1 v_2} + u_{w_1 w_2}) &= \underbrace{u_{v_1 v_2} u_{w_1 v_2}}_{0} + u_{v_1 v_2} u_{w_1 w_2} + u_{v_1 w_2} u_{w_1 v_2} + \underbrace{u_{v_1 w_2} u_{w_1 w_2}}_{0} \\ &= u_{v_1 v_2} u_{w_1 w_2} + u_{v_1 w_2} u_{w_1 v_2}. \end{align*} $$

On the other hand, Lemma 4.4 yields

$$ \begin{align*} (u_{v_1 v_2} + u_{v_1 w_2})(u_{w_1 v_2} + u_{w_1 w_2}) &= \sum_{\substack{v_3 \in V \\ \{v_1, v_3\} \in E}} u_{\{v_1, v_3\}\{v_2, w_2\}} \sum_{\substack{v_4 \in V \\ \{w_1, v_4\} \in E}} u_{\{w_1, v_4\}\{v_2, w_2\}} \\ &= \sum_{\substack{v_3, v_4 \in V \\ \{v_1, v_3\}, \{w_1, v_4\} \in E}} \underbrace{u_{\{v_1, v_3\}\{v_2, w_2\}} u_{\{w_1, v_4\}\{v_2, w_2\}}}_{\delta_{\{v_1, v_3\}\{w_1, v_4\}} u_{\{v_1, v_3\}\{v_2, w_2\}} } \\ &= u_{\{v_1, w_1\}\{v_2, w_2\}}, \end{align*} $$

where $\{v_1, v_3\} = \{w_1, v_4\}$ and $v_1 \neq w_1$ implies $v_3 = w_1$ in the last step. Thus,

$$\begin{align*}u_{\{v_1, w_1\}\{v_2, w_2\}} = (u_{v_1 v_2} + u_{v_1 w_2})(u_{w_1 v_2} + u_{w_1 w_2}) = u_{v_1 v_2} u_{w_1 w_2} + u_{v_1 w_2} u_{w_1 v_2}. \end{align*}$$

Similarly, we compute

$$\begin{align*}(u_{v_1 v_2} + u_{w_1 v_2}) (u_{v_1 w_2} + u_{w_1 w_2}) = u_{v_1 v_2} u_{w_1 w_2} + u_{w_1 v_2} u_{v_1 w_2} \end{align*}$$

and

$$ \begin{align*} (u_{v_1 v_2} + u_{w_1 v_2}) (u_{v_1 w_2} + u_{w_1 w_2}) &= \sum_{\substack{v_3, v_4 \in V \\ \{v_1, v_3\}, \{v_2, v_4\} \in E}} \underbrace{u_{\{v_1, w_1\} \{v_2, v_3\} } u_{\{v_1, w_1\} \{w_2, v_4\}}}_{\delta_{\{v_2, v_3\} \{w_2, v_4\}} u_{\{v_1, w_1\} \{v_2, v_3\} }} \\ &= u_{\{v_1, w_1\} \{v_2, w_2 \}} \end{align*} $$

using $v_2 \neq w_2$ and the other part of Lemma 4.4. Hence,

$$\begin{align*}u_{\{v_1, w_1\} \{v_2, w_2 \}} = u_{v_1 v_2} u_{w_1 w_2} + u_{w_1 v_2} u_{v_1 w_2} \quad \forall \{v_1, w_1\}, \{v_2, w_2\} \in E. \end{align*}$$

Proof Let $v \in V$ . By summing both sides of ${(A_s u_E)}_{ve} = {(u_V A_s)}_{ve}$ over all $e \in E$ , we obtain

$$ \begin{align*} \sum_{e \in E} \sum_{f \in E} {(A_s)}_{vf} u_{fe} &= \sum_{f \in E} {(A_s)}_{vf} \underbrace{\bigg(\sum_{e \in E} u_{fe}\bigg)}_{1} = \sum_{f \in E} {(A_s)}_{vf} = N_s(v), \\ \sum_{e \in E} \sum_{w \in V} u_{vw} {(A_s)}_{we} &= \sum_{w \in V} u_{vw} \bigg( \sum_{e \in E} {(A_s)}_{we} \bigg) = \sum_{w \in V} u_{vw} \cdot N_s(w). \end{align*} $$

Thus, $N_s(v) = \sum _{w \in V} u_{vw} \cdot N_s(w)$ , which implies

$$\begin{align*}N_s(v) \cdot u_{vw} = u_{vw} \cdot N_s(v) = \sum_{x \in V} \underbrace{u_{vw} u_{vx}}_{\delta_{wx} u_{vw}} \cdot N_s(x) = u_{vw} \cdot N_s(w) \quad \forall v,w \in V. \end{align*}$$

In particular, if $N_s(v) \neq N_s(w)$ , then

$$\begin{align*}N_s(v) \cdot u_{vw} = u_{vw} \cdot N_s(w) \ \Longleftrightarrow \ (N_s(v) - N_s(w)) \cdot u_{vw} = 0 \ \Longleftrightarrow \ u_{vw} = 0. \end{align*}$$

The corresponding statement and proof for the range map can be obtained by replacing s with r.

Proof Denote by u the fundamental representation of $\operatorname {\mathrm {Aut}}^+(\Gamma )$ and by $\widehat {u}$ the fundamental representation of $\operatorname {\mathrm {Aut}}^+_{\operatorname {\mathrm {Bic}}}(\Gamma )$ . Further, define the elements

$$\begin{align*}\widehat{u}_{\{v_1, w_1\} \{v_2, w_2\}} := \widehat{u}_{v_1 v_2} \widehat{u}_{w_1 w_2} + \widehat{u}_{v_1 w_2} \widehat{u}_{w_1 v_2} \quad \forall \{v_1, w_1\}, \{v_2, w_2\} \in E, \end{align*}$$

and the matrices $\widehat {u}_V := {(\widehat {u}_{vw})}_{v,w\in V}$ and $\widehat {u}_E := {(\widehat {u}_{ef})}_{e,f\in E}$ . Note that $\widehat {u}_{\{v_1, w_1\} \{v_2, w_2\}}$ is well-defined, since $\widehat {u}_{v_1 v_2} \widehat {u}_{w_1 w_2} = \widehat {u}_{w_1 w_2} \widehat {u}_{v_1 v_2}$ by Definition 2.27.

First, we construct the unital $*$ -homomorphism $\varphi \colon C(\operatorname {\mathrm {Aut}}^+(\Gamma )) \to C(\operatorname {\mathrm {Aut}}^+_{\operatorname {\mathrm {Bic}}}(\Gamma ))$ given by $\varphi (u_{vw}) = \widehat {u}_{vw}$ and $\varphi (u_{ef}) = \widehat {u}_{ef}$ using the universal property of $C(\operatorname {\mathrm {Aut}}^+(\Gamma ))$ . Thus, we have to show that $\widehat {u}_V$ and $\widehat {u}_E$ are magic unitaries that are intertwined by $A_s = A_r$ . By Definition 2.27, the matrix $\widehat {u}_V$ is a magic unitary and we can compute that

$$ \begin{align*} \widehat{u}_{\{v_1, w_1\}, \{v_2, w_2\}}^* &= {(\widehat{u}_{v_1 v_2} \widehat{u}_{w_1 w_2})}^* + {(\widehat{u}_{v_1 w_2} \widehat{u}_{w_1 v_2})}^* = \widehat{u}_{w_1 w_2} \widehat{u}_{v_1 v_2} + \widehat{u}_{w_1 v_2} \widehat{u}_{v_1 w_2} \\ &= \widehat{u}_{v_1 v_2} \widehat{u}_{w_1 w_2} + \widehat{u}_{v_1 w_2} \widehat{u}_{w_1 v_2} = \widehat{u}_{\{v_1, w_1\}, \{v_2, w_2\}} \end{align*} $$

for all $\{v_1, w_1\}, \{v_2, w_2\} \in E$ . Similarly, we compute

$$ \begin{align*} \widehat{u}_{\{v_1, w_1\}\{v_2, w_2\}}^2 &= (\widehat{u}_{v_1 v_2} \widehat{u}_{w_1 w_2} + \widehat{u}_{v_1 w_2} \widehat{u}_{w_1 v_2}) (\widehat{u}_{v_1 v_2} \widehat{u}_{w_1 w_2} + \widehat{u}_{v_1 w_2} \widehat{u}_{w_1 v_2}) \\ &= \widehat{u}_{v_1 v_2} \widehat{u}_{w_1 w_2} \widehat{u}_{v_1 v_2} \widehat{u}_{w_1 w_2} + \widehat{u}_{v_1 w_2} \widehat{u}_{w_1 v_2} \widehat{u}_{v_1 w_2} \widehat{u}_{w_1 v_2} \\ &= \widehat{u}_{v_1 v_2}^2 \widehat{u}_{w_1 w_2}^2 + \widehat{u}_{v_1 w_2}^2 \widehat{u}_{w_1 v_2}^2 \\ &= \widehat{u}_{v_1 v_2} \widehat{u}_{w_1 w_2} + \widehat{u}_{v_1 w_2} \widehat{u}_{w_1 v_2} \\ &= \widehat{u}_{\{v_1, w_1\}\{v_2, w_2\}} \end{align*} $$

for all $\{v_1, w_1\}, \{v_2, w_2\} \in E$ , where we additionally used the fact that $\widehat {u}_{w_1 w_2} \widehat {u}_{v_1 w_2} = 0$ and $\widehat {u}_{w_1 v_2} \widehat {u}_{v_1 v_2} = 0$ , since $v_1 \neq w_1$ . Next, we want to show that the rows and columns of $\widehat {u}_E$ sum up to $1$ . Observe that

$$\begin{align*}\sum_{\{v_1, w_1\} \in E} \widehat{u}_{v_1 v_2} \widehat{u}_{w_1 w_2} = \frac{1}{2} \sum_{\substack{v_1, w_1 \in V \\ \{v_1, w_1\} \in V}} \widehat{u}_{v_1 v_2} \widehat{u}_{w_1 w_2} = \frac{1}{2} \quad \forall \{v_2, w_2\} \in E \end{align*}$$

by Proposition 2.29. Therefore,

$$\begin{align*}\sum_{\{v_1, w_1\} \in E} \widehat{u}_{\{v_1, w_1\} \{v_2, w_2\}} = \sum_{\{v_1, w_1\} \in E} \widehat{u}_{v_1 v_2} \widehat{u}_{w_1 w_2} + \sum_{\{v_1, w_1\} \in E} \widehat{u}_{v_1 w_2} \widehat{u}_{w_1 v_2} = \frac{1}{2} + \frac{1}{2} = 1 \end{align*}$$

for all $\{v_2, w_2\} \in E$ . Similarly, we have

$$\begin{align*}\sum_{\{v_2, w_2\} \in E} \widehat{u}_{\{v_1, w_1\} \{v_2, w_2\}} = \frac{1}{2} + \frac{1}{2} = 1 \end{align*}$$

for all $\{v_1, w_1\} \in E$ . Hence, $\widehat {u}_E$ is a magic unitary. It remains to show that $A_s \widehat {u}_E = \widehat {u}_V A_s$ . But this follows from Lemma 4.4 and $A_{\Gamma } \widehat {u}_V = \widehat {u}_V A_{\Gamma }$ , because

$$ \begin{align*} \sum_{\substack{v_2 \in V \\ \{v_0, v_2\} \in E}} \widehat{u}_{\{v_0, v_2\}\{v_1, w_1\}} &= \sum_{\substack{v_2 \in V \\ \{v_0, v_2\} \in E}} \left( \widehat{u}_{v_0 v_1} \widehat{u}_{v_2 w_1} + \widehat{u}_{v_0 w_1} \widehat{u}_{v_2 v_1} \right) \\ &= \widehat{u}_{v_0 v_1} \sum_{\substack{v_2 \in V \\ \{v_0, v_2\} \in E}} \widehat{u}_{v_2 w_1} + \widehat{u}_{v_0 w_1} \sum_{\substack{v_2 \in V \\ \{v_0, v_2\} \in E}} \widehat{u}_{v_2 v_1} \\ &= \widehat{u}_{v_0 v_1} \sum_{v_2 \in V} {(A_{\Gamma})}_{v_0 v_2 } \widehat{u}_{v_2 w_1} + \widehat{u}_{v_0 w_1} \sum_{v_2 \in V} {(A_{\Gamma})}_{v_0 v_2} \widehat{u}_{v_2 v_1} \\ &= \widehat{u}_{v_0 v_1} \sum_{v_2 \in V} \widehat{u}_{v_0 v_2 } {(A_{\Gamma})}_{v_2 w_1} + \widehat{u}_{v_0 w_1} \sum_{v_2 \in V} \widehat{u}_{v_0 v_2} {(A_{\Gamma})}_{v_2 v_1} \\ &= \sum_{v_2 \in V} \underbrace{\widehat{u}_{v_0 v_1} \widehat{u}_{v_0 v_2 }}_{\delta_{v_1 v_2} \widehat{u}_{v_0 v_1}} {(A_{\Gamma})}_{v_2 w_1} + \sum_{v_2 \in V} \underbrace{\widehat{u}_{v_0 w_1} \widehat{u}_{v_0 v_2}}_{\delta_{w_1 v_2} \widehat{u}_{v_0 w_1}} {(A_{\Gamma})}_{v_2 v_1} \\ &= \widehat{u}_{v_0 v_1} + \widehat{u}_{v_0 w_1} \end{align*} $$

for all $v_0 \in V$ and $\{v_1, w_1\} \in E$ . Hence, the map $\varphi $ exists.

Next, we construct the inverse map $ \psi \colon C(\operatorname {\mathrm {Aut}}^+_{\operatorname {\mathrm {Bic}}}(\Gamma )) \to C(\operatorname {\mathrm {Aut}}^+(\Gamma ))$ given by $\psi (\widehat {u}_{vw}) = u_{vw}$ using the universal property of $C(\operatorname {\mathrm {Aut}}^+_{\operatorname {\mathrm {Bic}}}(\Gamma ))$ . Hence, we have to show that $u_V$ satisfies the relations from Definition 2.27. By Definition 3.4, $u_V$ is a magic unitary. To show that $A_{\Gamma } u_V = u_V A_{\Gamma }$ , observe that

$$\begin{align*}{(A_s A_s^*)}_{v w} = \sum_{e \in E} {(A_s)}_{v e} {(A_s)}_{w e} = \left\lvert\{ e \in E ~|~ v \in s(e) \ \land \ w \in s(e) \}\right\rvert \quad \forall v, w \in V. \end{align*}$$

Since $\Gamma $ is a simple graph, each $e \in E$ contains exactly two elements so that

$$\begin{align*}{(A_s A_s^*)}_{v w} = \begin{cases} 0 & \text{if } v \neq w, \{v, w\} \notin E, \\ 1 & \text{if } v \neq w, \{v, w\} \in E, \\ N_s(v) & \text{if } v = w, \\ \end{cases} \end{align*}$$

where $N_s(v) := \left \lvert \{ e \in E ~|~ v \in s(e) \}\right \rvert $ as in Proposition 4.6. Hence, we can write $A_s A_s^* = A_{\Gamma } + T$ with $T \in \mathbb {C}^{V \times V}$ defined by

$$\begin{align*}T_{vw} := \begin{cases} 0 & \text{if }v \neq w, \\ N_s(v) & \text{if }v = w, \end{cases} \quad \forall v, w \in V. \end{align*}$$

Proposition 4.6 shows that $T u_V = u_V T$ , which implies

$$\begin{align*}A_{\Gamma} u_V = A_s A_s^* u_V - T u_V = A_s u_E A_s^* - u_V T = u_V A_s A_s^* - u_V T = u_V A_{\Gamma}. \end{align*}$$

Finally, we have to show that $u_{v_1 v_2} u_{w_1 w_2} = u_{w_1 w_2} u_{v_1 v_2}$ for all $\{v_1, w_1\}, \{v_2, w_2\} \in E$ . But this follows directly from Lemma 4.5, since

$$ \begin{align*} u_{v_1 v_2} u_{w_1 w_2} &= u_{\{v_1, w_1\}\{v_2,w_2\}} - u_{v_1 w_2} u_{w_1 v_2} = u_{\{w_1, v_1\}\{w_2,v_2\}} - u_{v_1 w_2} u_{w_1 v_2} \\ &= u_{w_1 w_2} u_{v_1 v_2} + u_{v_1 w_2} u_{w_1 v_2} - u_{v_1 w_2} u_{w_1 v_2} = u_{w_1 w_2} u_{v_1 v_2}. \end{align*} $$

Thus, the $*$ -homomorphism $\psi $ exists. By the definitions of $\varphi $ and $\psi $ , we have

$$\begin{align*}u_{vw} \ \longleftrightarrow \ \widehat{u}_{vw}, \qquad u_{\{v_1, w_1\}\{v_2,w_2\}} \ \longleftrightarrow \ \widehat{u}_{v_1 v_2} \widehat{u}_{w_1 w_2} + \widehat{u}_{v_1 w_2} \widehat{u}_{w_1 v_2} \end{align*}$$

for all $v, w \in V$ and $\{v_1, w_1\}, \{v_2, w_2\} \in E$ , which shows that both maps are indeed inverse. Further, one directly verifies that $\psi $ is a morphism of compact quantum groups. Hence, $\psi $ is an isomorphism of compact quantum groups, which shows $\operatorname {\mathrm {Aut}}^+(\Gamma ) = \operatorname {\mathrm {Aut}}^+_{\operatorname {\mathrm {Bic}}}(\Gamma )$ .

Proof First, consider the equation $A_s u_E = u_V A_s$ , which is equivalent to $A_s^* u_V = u_E A_s^*$ by Proposition 2.24. A direct computation yields

$$\begin{align*}{(u_E A^*_s)}_{ev} = \sum_{\substack{f \in E \\ s(f) = v}} u_{ef}, \quad {(A^*_s u_V)}_{ev} = \sum_{\substack{w \in V \\ s(e) = w}} u_{wv} = u_{s(e) v} \qquad \forall v \in V, \, e \in E, \end{align*}$$

such that $A_s^* u_V = u_E A_s^*$ is equivalent to $\sum _{\substack {f \in E \\ s(f) = v}} u_{ef} = u_{s(e) v}$ . The statement for the range map r can be obtained by replacing s with r in the previous computation.

Proof First, note that at least one case in Definition 4.9 applies to each vertex v, since $\Gamma $ has no isolated vertices. Further, each case does not depend on the edge e by Relation 2 in Definition 2.30. To show that overlapping cases are well-defined, assume $v, w \in V$ such that v is neither a source nor a sink. If w is neither a source nor a sink, then both cases agree by Relation 4. On the other hand, if w is a source or a sink, then both cases yield ${u}_{vw} = 0$ , since the sum is empty in one case, while each ${u}_{ef} = 0$ by Relation 3 in the other case. Thus, the elements ${u}_{vw}$ are well-defined.

Next, we show that the matrix $u_V := {(u_{vw})}_{v,w \in V}$ is a magic unitary. Let $v, w \in V$ and assume without loss of generality that there exists an edge $e \in E$ with $v = s(e)$ . Using Lemma 4.8 and the fact that ${(u_{ef})}_{e,f \in E}$ is a magic unitary, we compute

$$ \begin{align*} u_{vw}^* &=\sum_{\substack{f \in E \\ s(f) = w}} u_{e f}^* =\sum_{\substack{f \in E \\ s(f) = w}} u_{e f} = u_{vw}, \\ u_{vw}^2 &= \sum_{\substack{f_1 \in E \\ s(f_1) = w}} \sum_{\substack{f_2 \in E \\ s(f_2) = w}} \underbrace{u_{e f_1} u_{e f_2}}_{\delta_{f_1 f_2} u_{e f_1}} = \sum_{\substack{f \in E \\ s(f) = w}} u_{e f} = u_{vw} \end{align*} $$

and

$$\begin{align*}\sum_{w \in V} u_{vw} = \sum_{w \in V} \sum_{\substack{f \in E \\ s(f) = w}} u_{e f} = \sum_{f \in E} \underbrace{\sum_{\substack{w \in V \\ s(f) = w}} u_{e f}}_{u_{e f}} = \sum_{f \in E} u_{e f} = 1. \end{align*}$$

To show that the rows of $u_V$ sum to $1$ , we use an argument from the proof of [Reference Wang34, Theorem 3.1]. By our previous computation, we have

$$\begin{align*}{(u_V u_V^*)}_{vw} = \sum_{x \in V} \underbrace{u_{vx} u_{wx}^*}_{\delta_{vw} u_{vx}} = \delta_{vw} \sum_{x \in V} u_{vx} = \delta_{vw} \quad \forall v, w \in V. \end{align*}$$

Hence, $u_V$ is right-invertible with $u_V u_V^* = 1$ . If we show that $u_V$ is a representation of $\operatorname {\mathrm {Aut}}^+_{\operatorname {\mathrm {GH}},\operatorname {\mathrm {Bic}}}(\Gamma )$ , that is, $\Delta (u_{vw}) = \sum _{x \in V} u_{vx} \otimes u_{xw}$ , then [Reference Woronowicz36, Proposition 3.2] implies that $u_V$ is also left-invertibe, that is, $u_V^* u_V = 1$ . Thus,

$$\begin{align*}1 = {(u_V^* u_V)}_{vv} = \sum_{w \in V} u_{wv}^* u_{wv} = \sum_{w \in V} u_{wv} \quad \forall v \in V. \end{align*}$$

Therefore, it remains to show that $u_V$ is a representation of $\operatorname {\mathrm {Aut}}^+_{\operatorname {\mathrm {GH}},\operatorname {\mathrm {Bic}}}(\Gamma )$ . Again, let $v, w \in V$ and assume without loss of generality that there exists an edge $e \in E$ with $v = s(e)$ . Then

$$ \begin{align*} \Delta(u_{vw}) &= \sum_{\substack{f \in E \\ s(f) = w}} \Delta(u_{ef}) = \sum_{\substack{f \in E \\ s(f) = w}} \sum_{g \in E} u_{eg} \otimes u_{gf} = \sum_{g \in E} u_{eg} \otimes \bigg( \sum_{\substack{f \in E \\ s(f) = w}} u_{gf} \bigg) \\ &= \sum_{g \in E} u_{eg} \otimes u_{s(g) w,} \end{align*} $$

and on the other hand,

$$ \begin{align*} \sum_{x \in V} u_{vx} \otimes u_{xw} = \sum_{x \in V} \sum_{\substack{f \in E \\ s(f) = x}} u_{e f} \otimes u_{xw} = \sum_{f \in E} \sum_{\substack{x \in V \\ s(f) = x}} u_{e f} \otimes u_{xw} = \sum_{\substack{f \in E}} u_{e f} \otimes u_{s(f) w}. \end{align*} $$

Thus, $\Delta (u_{vw}) = \sum _{x \in V} u_{vx} \otimes u_{xw}$ for all $v, w \in V$ .

Proof Let $\Gamma := (V, E)$ . Denote by u the fundamental representation of $\operatorname {\mathrm {Aut}}^+(\Gamma )$ and by $\widehat {u}$ the fundamental representation of $\operatorname {\mathrm {Aut}}^+_{\operatorname {\mathrm {GH}},\operatorname {\mathrm {Bic}}}(\Gamma )$ . Further, define the elements

$$\begin{align*}\widehat{u}_{vw} := \begin{cases} \sum_{\substack{f \in E \\ s(f) = w}} \widehat{u}_{e f} & \text{if there exists } e \in E\ \text{with } s(e) = v, \\ \sum_{\substack{f \in E \\ r(f) = w}} \widehat{u}_{e f} & \text{if there exists } e \in E\ \text{with } r(e) = v, \end{cases} \end{align*}$$

for all $v, w \in V$ as in Definition 4.9.

We begin by constructing the unital $*$ -homomorphism $\varphi \colon C(\operatorname {\mathrm {Aut}}^+(\Gamma )) \to C(\operatorname {\mathrm {Aut}}^+_{\operatorname {\mathrm {GH}},\operatorname {\mathrm {Bic}}}(\Gamma ))$ given by $\varphi (u_{vw}) = \widehat {u}_{vw}$ and $\varphi (u_{ef}) = \widehat {u}_{ef}$ using the universal property of $C(\operatorname {\mathrm {Aut}}^+(\Gamma ))$ . Hence, we have to show that the matrices $\widehat {u}_V := {(\widehat {u}_{vw})}_{v,w\in V}$ and $\widehat {u}_E := {(\widehat {u}_{ef})}_{e,f\in E}$ satisfy the relations of Definition 3.4. The matrix $\widehat {u}_E$ is a magic unitary by Relation 1 of Definition 2.30 and $\widehat {u}_V$ is a magic unitary by Lemma 4.10. Next, consider the relation $A_s \widehat {u}_E = \widehat {u}_V A_s$ , which is equivalent to $A_s^* \widehat {u}_V = \widehat {u}_E A_s^*$ by Proposition 2.24. However, this relation follows directly, since

$$\begin{align*}{(A_s^* \widehat{u}_V)}_{ev} = \sum_{w \in V} {(A_s^*)}_{ew} \widehat{u}_{wv} = \widehat{u}_{s(e) v} = \sum_{\substack{f \in E \\ s(f) = v}} \widehat{u}_{ef} = \sum_{f \in E} \widehat{u}_{ef} {(A_s^*)}_{fv} = {(\widehat{u}_E A^*_s)}_{ev} \end{align*}$$

for all $e \in E$ and $v \in V$ . Similarly, one shows $A_r \widehat {u}_E = \widehat {u}_V A_r$ , such that the $*$ -homomorphism $\varphi $ exists.

Next, we construct the inverse $*$ -homomorphism $\psi \colon C(\operatorname {\mathrm {Aut}}^+_{\operatorname {\mathrm {GH}},\operatorname {\mathrm {Bic}}}(\Gamma )) \to C(\operatorname {\mathrm {Aut}}^+(\Gamma ))$ given by $\psi (\widehat {u}_{ef}) = u_{ef}$ using the universal property of $C(\operatorname {\mathrm {Aut}}^+_{\operatorname {\mathrm {GH}},\operatorname {\mathrm {Bic}}}(\Gamma ))$ . Thus, we have to show that the entries of $u_E$ satisfy the relations in Definition 2.30. By Definition 3.4, $u_E$ is a magic unitary. Further, we can use Lemma 4.8 to compute

$$ \begin{align*} \sum_{\substack{f \in E \\ s(f) = v}} u_{e_1 f} &= u_{s(e_1) v} = u_{s(e_2) v} = \sum_{\substack{f \in E \\ s(f) = v}} u_{e_2 f}, \\ \sum_{\substack{f \in E \\ r(f) = v}} u_{e_1 f} &= u_{r(e_1) v} = u_{r(e_2) v} = \sum_{\substack{f \in E \\ r(f) = v}} u_{e_2 f} \end{align*} $$

for all $v \in V$ and $e_1, e_2 \in E$ with $s(e_1) = s(e_2)$ or $r(e_1) = r(e_2),$ respectively. Thus, Relation 2 is satisfied. Next, consider Relation 3 and let $e, f \in E$ . If $s(e)$ is neither a source or sink and $s(f)$ is a source, then $N_{r}(s(e))> 0$ and $N_{r}(s(f)) = 0$ in the notation of Proposition 4.6. Therefore, Proposition 4.6 implies that $u_{s(e)s(f)} = 0$ , such that

$$\begin{align*}0 = u_{s(e)s(f)} u_{ef} = \sum_{\substack{ g \in E \\ s(g) = s(f)}} \underbrace{u_{eg} u_{ef}}_{\delta_{gf} u_{ef}} = u_{ef}. \end{align*}$$

Similarly, one shows $u_{s(e)s(f)} = 0$ if $r(f)$ is a sink. Hence, Relation 3 is satisfied. Finally, Lemma 4.8 implies

$$\begin{align*}\sum_{\substack{f \in E \\ s(f) = v}} u_{e_1 f} = u_{s(e_1) v} = u_{r(e_2) v} = \sum_{\substack{f \in E \\ r(f) = v}} u_{e_2 f} \end{align*}$$

for all $v \in V$ and $e_1, e_2 \in E$ with $s(e_1) = r(e_2)$ . Thus, Relation 4 holds and the $*$ -homomorphism $\varphi $ exist.

The $*$ -homomorphisms $\varphi $ and $\phi $ are indeed inverse, since

$$\begin{align*}u_{ef} \ \longleftrightarrow \ \widehat{u}_{ef} \quad \forall e, f \in E, \qquad u_{vw} \ \longleftrightarrow \ \widehat{u}_{vw} \quad \forall v, w \in V \end{align*}$$

by Lemma 4.8 and Definition 4.9. Further, a direct computation shows that $\psi $ is a morphism of compact quantum groups. Hence, $\operatorname {\mathrm {Aut}}^+(\Gamma ) = \operatorname {\mathrm {Aut}}^+_{\operatorname {\mathrm {GH}},\operatorname {\mathrm {Bic}}}(\Gamma )$ .

Proof We use the universal property of $C^*(\Gamma )$ to construct the map $\alpha $ from the statement. Thus, we have to show that $\alpha (p_v)$ are orthogonal projections, $\alpha (s_e)$ are partial isometries, and both satisfy the relations from Definition 2.33. Recall the magic unitary relations of $u_V$ and $u_E$ and the intertwiner relations of $A_s$ and $A_r$ from Remark 3.6. Then, we can show that $\alpha (p_v)$ are orthogonal projections by computing

$$ \begin{align*} \alpha(p_{v_1}) \alpha(p_{v_2}) &= \sum_{w_1, w_2 \in V} \underbrace{p_{w_1} p_{w_2}}_{\delta_{w_1 w_2} p_{w_1}} \otimes u_{w_1 v_1} u_{w_2 v_2} \\ &= \sum_{w \in V} p_w \otimes \underbrace{u_{w v_1} u_{w v_2}}_{\delta_{v_1 v_2} u_{w v_1}} = \delta_{v_1 v_2} \sum_{w \in V} p_w \otimes u_{w v_1} = \delta_{v_1 v_2} \, \alpha(p_{v_1}) \end{align*} $$

for all $v_1, v_2 \in V$ and

$$ \begin{align*} {\alpha(p_v)}^* = \sum_{w \in V} p_w^* \otimes u_{w v}^* = \sum_{w \in V} p_w \otimes u_{w v} = \alpha(p_v) \end{align*} $$

for all $v \in V$ . Similarly, we show that $\alpha (s_e)$ are partial isometries by computing

$$ \begin{align*} \alpha(s_e){\alpha(s_e)}^*\alpha(s_e) &= \sum_{f_1, f_2, f_3 \in E} s_{f_1} s_{f_2} s_{f_3} \otimes \underbrace{u_{f_1 e} u_{f_2 e}^* u_{f_3 e}}_{\delta_{f_1 f_2} \delta_{f_1 f_3} u_{f_1 e}} \\ &= \sum_{f \in E} \underbrace{s_f s_f^* s_f}_{s_f} \otimes u_{f e} = \sum_{f \in E} s_f \otimes u_{f e} = \alpha(s_e) \end{align*} $$

for all $e \in E$ . Next, consider Relation 1 from Definition 2.33, which states that $s_e^* s_f = \delta _{ef} \sum _{\substack {v \in V \\ v \in r(e)}} p_v$ for all $e,f \in E$ . When applying $\alpha $ to the left side, we obtain

$$ \begin{align*} {\alpha(s_e)}^* \alpha(s_f) &= \sum_{g_1, g_2 \in E} \underbrace{s_{g_1}^* s_{g_2}}_{\text{Rel. 1}} \otimes u_{g_1 e}^* u_{g_2 f} \\ &= \sum_{g \in E} \sum_{\substack{v \in V \\ v \in r(g)}} p_v \otimes \underbrace{u_{g e} u_{g f}}_{\delta_{e f} u_{g e}} = \delta_{ef} \sum_{v \in V} \sum_{\substack{g \in E \\ v \in r(g)}} p_v \otimes u_{g e}. \end{align*} $$

Using Remark 3.6, we have $\sum _{\substack {g \in E \\ v \in r(g)}} u_{ge} = \sum _{\substack {w \in V \\ w \in r(e)}} u_{vw}$ for all $v \in V$ and $e \in E$ , which can be used to further rewrite ${\alpha (s_e)}^* \alpha (s_f)$ to

$$ \begin{align*} {\alpha(s_e)}^* \alpha(s_f) &= \delta_{e f} \sum_{v \in V} \sum_{\substack{w \in V \\ w \in r(e)}} p_v \otimes u_{v w} = \delta_{e f} \sum_{\substack{w \in V \\ w \in r(e)}} \sum_{v \in V} p_v \otimes u_{v w} \\ &= \delta_{e f} \sum_{\substack{w \in V \\ w \in r(e)}} \alpha(p_w). \end{align*} $$

Thus, Relation 1 is satisfied. Next, consider Relation 2, which is given by $s_e s_e^* \leq \sum _{\substack {v \in V \\ v \in s(e)}} p_v$ for all $e \in E$ . As before, we apply $\alpha $ to the left side and compute

$$\begin{align*}\alpha(s_e) {\alpha(s_e)}^* &= \sum_{f_1, f_2 \in E} s_{f_1} s_{f_2}^* \otimes \underbrace{u_{f_1 e} u_{f_2 e}^*}_{\delta_{f_1 f_2} u_{f_1 e}} = \sum_{f \in E} s_{f} s_{f}^* \otimes u_{f e}. \end{align*}$$

Since each $u_{fe} \geq 0$ , we can use Relation 2 to obtain

$$\begin{align*}\alpha(s_e) {\alpha(s_e)}^* \leq \sum_{f \in E} \bigg( \sum_{\substack{v \in V \\ v \in s(f)}} p_v \bigg) \otimes u_{f e} = \sum_{v \in V } \sum_{\substack{f \in E \\ v \in s(f)}} p_v \otimes u_{f e}. \end{align*}$$

By Remark 3.6, we have $\sum _{\substack {f \in E \\ v \in s(f)}} u_{f e} = \sum _{\substack {w \in V \\ w \in s(e)}} u_{w v}$ for all $v \in V$ and $e \in E$ , such that

$$\begin{align*}\alpha(s_e) {\alpha(s_e)}^* \leq \sum_{v \in V } \sum_{\substack{w \in V \\ w \in s(e)}} p_v \otimes u_{vw} = \sum_{\substack{w \in V \\ w \in s(e)}} \sum_{v \in V } p_v \otimes u_{vw} = \sum_{\substack{w \in V \\ w \in s(e)}} \alpha(p_w). \end{align*}$$

Hence, Relation 2 is satisfied. Finally, consider Relation 3, which states that $p_v \leq \sum _{\substack {e \in E \\ v \in s(e)}} s_e s_e^*$ for all $v \in V$ that are not a sink. Let $v \in V$ be a vertex that is not a sink. By Proposition 4.6, we have $u_{vw} = 0$ for all $w \in V$ that are a sink, since $N_s(v)> 0$ if v is not a sink and $N_s(w) = 0$ if w is a sink. Therefore,

$$\begin{align*}\alpha(p_v) = \sum_{w \in V} p_w \otimes u_{w v} = \sum_{\substack{w \in V \\ w \neq \text{sink }}} p_w \otimes u_{w v}. \end{align*}$$

Since $u_{vw} \geq 0$ , Relation 3 yields

$$\begin{align*}\alpha(p_v) \leq \sum_{\substack{w \in V \\ {w \neq sink}}} \bigg( \sum_{\substack{e \in E \\ w \in s(e)}} s_e s_e^* \bigg) \otimes u_{w v} = \sum_{e \in E} \sum_{\substack{w \in V \\ w \in s(e)}} s_e s_e^* \otimes u_{w v}, \end{align*}$$

where we used again the fact that $u_{v w} = 0$ if w is a sink. On the other hand, we have

$$ \begin{align*} \sum_{\substack{f \in E \\ v \in s(f)}} \alpha(s_f) {\alpha(s_f)}^* = \sum_{e_1, e_2 \in E} \sum_{\substack{f \in E \\ v \in s(f)}} s_{e_1} s_{e_2}^* \otimes \underbrace{u_{e_1 f} u_{e_2 f}^*}_{\delta_{e_1 e_2} u_{e_1 f}} = \sum_{e \in E} \sum_{\substack{f \in E \\ v \in s(f)}} s_e s_e^* \otimes u_{e f}. \end{align*} $$

By Remark 3.6, we have $ \sum _{\substack {w \in V \\ w \in s(e)}} u_{w v} = \sum _{\substack {f \in E \\ v \in s(f)}} u_{e f} $ for all $v \in V$ and $e \in E$ , which implies

$$ \begin{align*} \alpha(p_v) \leq \sum_{e \in E} \sum_{\substack{w \in V \\ w \in s(e)}} s_e s_e^* \otimes u_{w v} = \sum_{e \in E} \sum_{\substack{f \in E \\ v \in s(f)}} s_e s_e^* \otimes u_{e f} = \sum_{\substack{f \in E \\ v \in s(f)}} \alpha(s_f) {\alpha(s_f)}^*. \end{align*} $$

Proof Let $\Gamma := (V, E)$ and define $\mathcal {B} \subseteq C^*(\Gamma )$ as the $*$ -subalgebra generated by $p_v$ for all $v \in V$ and $p_e$ for all $e \in E$ . Then $\mathcal {B}$ is dense in $C^*(\Gamma )$ and we have $\alpha (\mathcal {B}) \subseteq \mathcal {B} \otimes \mathcal {O}(\operatorname {\mathrm {Aut}}^+(\Gamma ))$ by the definition of $\alpha $ . Next, let $v \in V$ . Then

$$\begin{align*}\sum_{w \in V} \alpha(p_w) \otimes u_{wv} = \sum_{w_1, w_2 \in V} p_{w_2} \otimes u_{w_2 w_1} \otimes u_{w_1 v} = \sum_{w_2 \in V} p_{w_2} \otimes \Delta(u_{w_2 v}) \end{align*}$$

implies $(\alpha \otimes \operatorname {\mathrm {id}})(\alpha (p_v)) = (\operatorname {\mathrm {id}} \otimes \Delta )(\alpha (p_v))$ . Further, we compute

$$\begin{align*}(\operatorname{\mathrm{id}} \otimes \varepsilon)(\alpha(p_v)) = (\operatorname{\mathrm{id}} \otimes \varepsilon)\bigg(\sum_{w \in V} p_{w} \otimes u_{wv} \bigg) = \sum_{w \in V} p_w \cdot \delta_{wv} = p_v. \end{align*}$$

The previous computations also show that $(\alpha \otimes \operatorname {\mathrm {id}})(\alpha (s_e)) = (\operatorname {\mathrm {id}} \otimes \Delta )(\alpha (s_e))$ and $(\operatorname {\mathrm {id}} \otimes \varepsilon )(\alpha (s_e)) = s_e$ for all $e \in E$ by replacing $p_v$ with $s_e$ . Hence, $(\operatorname {\mathrm {id}} \otimes \alpha ) \circ \alpha = (\Delta \otimes \operatorname {\mathrm {id}}) \circ \alpha $ and $(\varepsilon \otimes \operatorname {\mathrm {id}}) \circ \alpha |_{\mathcal {B}} = \operatorname {\mathrm {id}}$ . Thus, $\alpha $ defines an action of $\operatorname {\mathrm {Aut}}^+(\Gamma )$ on $C^*(\Gamma )$ . To show that $\alpha $ is faithful, assume that there exists a quotient G of $\operatorname {\mathrm {Aut}}^+(\Gamma )$ such that $\alpha |_{C(G)}$ is also an action on $C^*(\Gamma )$ . Then

$$ \begin{align*} \alpha|_{C(G)}(p_v) &= \sum_{w \in V} p_w \otimes u_{wv} \in C^*(\Gamma) \otimes C(G) \qquad \forall v \in V, \\ \alpha|_{C(G)}(s_e) &= \sum_{f \in E} s_f \otimes u_{fe} \in C^*(\Gamma) \otimes C(G) \qquad \forall e \in E. \end{align*} $$

The representation of $\alpha |_{C(G)}(p_v)$ with respect to $p_v$ is unique since the $p_v$ are linearly independent as orthogonal projections. Thus, $u_{wv} \in C(G)$ for all $v, w \in V$ . Similarly, $u_{ef} \in C(G)$ for all $e, f \in E$ , since the $s_e$ are linearly independent as partial isometries with orthogonal ranges. Hence, $C(G) = C(\operatorname {\mathrm {Aut}}^+(\Gamma ))$ , which shows that $\alpha $ is faithful.

Proof Denote by u the fundamental representation of $\operatorname {\mathrm {Aut}}^+(\Gamma )$ and by $\widehat {u}$ the fundamental representation of $\operatorname {\mathrm {Aut}}^+(\Gamma ')$ . By Propositions 3.12 and 3.4, we have

$$\begin{align*}\operatorname{\mathrm{Aut}}^+(\Gamma') = \operatorname{\mathrm{Aut}}^+\left({(\Gamma^*)}^{\operatorname{\mathrm{op}}}\right) = \operatorname{\mathrm{Aut}}^+(\Gamma^*) = \operatorname{\mathrm{Aut}}^+(\Gamma) \end{align*}$$

via the mapping $\widehat {u}_V \mapsto u_E$ and $\widehat {u}_E \mapsto u_V$ . The statement follows by applying this isomorphism to the action of $\operatorname {\mathrm {Aut}}^+(\Gamma ')$ on $C^*(\Gamma ')$ from Theorem 5.2.

Proof The proof that $u_V$ is a magic unitary is contained in the proof of [Reference Wang34, Theorem 3.1], since the elements $p_v$ are orthogonal projections that sum up to $1$ . For the second part of the statement, assume that $u_E$ is also a magic unitary and consider the relation $s_e^* s_e = \sum _{\substack {v \in V \\ v \in r(e)}} p_v$ from Definition 2.33. By applying $\alpha $ to the left side and using the relation, we obtain

$$\begin{align*}\alpha(s_e^* s_e) = \sum_{f_1, f_2 \in E} s_{f_1}^* s_{f_2} \otimes \underbrace{u_{f_1 e}^* u_{f_2 e}}_{\delta_{f_1 f_2} u_{f_1 e}} = \sum_{f \in E} s_f^* s_f \otimes u_{f e} = \sum_{f \in E} \sum_{\substack{v \in V \\ v \in r(f)}} p_v \otimes u_{f e}, \end{align*}$$

which can be rewritten as

$$\begin{align*}\alpha(s_e^* s_e) = \sum_{v \in V} \sum_{f \in E} {(A_r)}_{vf} \, p_v \otimes u_{f e} = \sum_{v \in V} p_v \otimes \bigg(\sum_{f \in E} {(A_r)}_{v f} \, u_{f e} \bigg). \end{align*}$$

Similarly, by applying $\alpha $ to the right side of the original equation, we obtain

$$ \begin{align*} \sum_{\substack{v \in V \\ v \in r(e)}} \alpha(p_v) &= \sum_{\substack{v \in V \\ v \in r(e)}} \sum_{w \in V} p_w \otimes u_{w v} \\ &= \sum_{v, w \in V} {(A_r)}_{v e} \, p_w \otimes u_{w v} = \sum_{w \in V} p_w \otimes \bigg( \sum_{v \in V} u_{w v} {(A_r)}_{v e} \bigg). \end{align*} $$

Thus,

$$\begin{align*}{(A_r u_E)}_{v e} = \sum_{f \in E} {(A_r)}_{v f} \, u_{f e} = \sum_{w \in V} u_{w v} {(A_r)}_{v e} = {(A_r u_V)}_{v e} \quad \forall v \in V \end{align*}$$

by using the fact that the elements $p_v$ are linearly independent as orthogonal projections, so that we can compare the terms in the previous sums. This shows $A_r u_E = u_V A_r$ .

Proof By applying the first part of Lemma 5.6 to $\alpha _1$ and $\alpha _2$ , we obtain that the matrices $u_V := {(u_{vw})}_{v,w \in V}$ and $u_E := {(u_{ef})}_{e,f\in E}$ are magic unitaries. The second part of Lemma 5.6 then yields $A_r u_E = u_V A_r$ and $A_{r'} u_V = u_E A_{r'}$ . But $A_{r'} = A_s^*$ , such that $A_s^* u_V = u_E A_s^*$ , which implies $A_s u_E = u_V A_s$ by Proposition 2.24. This shows that the elements $u_{vw}$ and $u_{ef}$ satisfy the relations of Definition 3.4. Hence, by the universal property of $C(\operatorname {\mathrm {Aut}}^+(\Gamma ))$ , there exists a unital $*$ -homomorphism $\varphi \colon C(\operatorname {\mathrm {Aut}}^+(\Gamma )) \to C(G)$ that maps the generators of $C(\operatorname {\mathrm {Aut}}^+(\Gamma ))$ to the entries of $u_V$ and $u_E$ . Next, we show that $\varphi $ is a morphism of compact quantum groups. Let $w \in V$ and observe that

$$ \begin{align*} (\alpha_1 \otimes \operatorname{\mathrm{id}})(\alpha_1(w)) &= \sum_{x \in V} \alpha_1(p_{x}) \otimes u_{x w} \\ &= \sum_{x \in V} \sum_{v \in V} p_{v} \otimes u_{v x} \otimes u_{x w} = \sum_{v \in V} p_{v} \otimes \bigg(\sum_{x \in V} u_{v x} \otimes u_{x w}\bigg), \end{align*} $$

which implies $(\operatorname {\mathrm {id}} \otimes \Delta )(\alpha _1(w)) = \sum _{v \in V} p_{v} \otimes \Delta (u_{v w})$ . Because $\alpha _1$ is an action, it satisfies $(\alpha _1 \otimes \operatorname {\mathrm {id}}) \circ \alpha _1 = (\operatorname {\mathrm {id}} \otimes \Delta ) \circ \alpha _1$ . Further, all $p_v$ are linearly independent as orthogonal projections, which implies $\Delta (u_{vw}) = \sum _{w \in V} u_{v x} \otimes u_{x w}$ for all $v, w \in V$ . Denote by $\widehat {u}$ the fundamental representation of $\operatorname {\mathrm {Aut}}^+(\Gamma )$ . Then the previous equation yields